what is the missing exponent that would make this equation true?

(k^(5)*k^(2))/(k^(x))=(k^(2))^(-2)
a. x=3
b. x=11
c. x=9
d. x=0
e. x=1
f. x=6

1 answer

The correct answer is:

a. x = 3

Explanation:
(k^5 * k^2) / (k^x) = k^(5 + 2) / k^(x) = k^7 / k^x = k^(7 - x)

(k^2)^(-2) = k^(2 * -2) = k^(-4)

Now, we need to set the two expressions equal to each other:

k^(7 - x) = k^(-4)

7 - x = -4
x = 7 + 4
x = 3

Therefore, the missing exponent that would make this equation true is x = 3.